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题目对θ∈[0,π/2],关于x的方程(x+3+sin 2θ)~2+(x+asin θ+acos θ)~2=1/8恒有解,求实数a的取值范围.分析本题有三个变量x,a,θ,已知的是方程有解,要求的是a的取值范围,根据题意,就是要从等式出发列出不等式,并消去变量x,θ.由于x是主元,我们可以考虑先消去x,再消去θ,最后求出a的取值范围.本题的突破口就是如何从方程出发消去x,列出关于a,θ的不等式.下面从四个角
The problem is always solved for θ∈ [0, π / 2] and the equation (x + 3 + sin 2θ) ~ 2 + (x + asin θ + acos θ) ~ 2 = 1/8 for x, Analysis of the problem There are three variables x, a, θ, known is the equation has a solution, the requirement is a range of values, according to the title, is to start from the equations listed inequalities, and eliminate variables x, θ. Since x is the principal, we can consider eliminating x, eliminating θ and finally finding the range of a. The breakthrough of this problem is how to eliminate x from the equation and list the inequalities about a and θ. Below from the four corners